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Functions Test 20

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Functions Test 20
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  • Question 1
    1 / -0
    The domain of the function, f(x)= 1[x]2[x]6f(x) = \displaystyle \dfrac {1}{\sqrt{[x]^2-[x]-6}}  is
    Solution
    Given : f(x)= 1[x]2[x]6f(x) = \displaystyle \frac {1}{\sqrt{[x]^2-[x]-6}}

    f(x)f(x) is real if [x]2[x]6R{\sqrt{[x]^2-[x]-6}}\in R 

    i.e.
     [x]2[x]6>0[x]^{ 2 }-[x]-6>0
        [x]23[x]+2[x]6>0\implies [x]^{ 2 }-3[x]+2[x]-6>0
        [x]([x]3)+2([x]3)>0\implies [x]([x]-3)+2([x]-3)>0
        ([x]3)([x]+2)>0\implies ([x]-3)([x]+2)>0
        [x]<2 \implies \left[ x \right] <-2 and [x]>3\left[ x \right] >3

    Therefore, x (,3][4,)x \in (-\infty , -3] \cup [4, \infty )
  • Question 2
    1 / -0
    The domain of the function ln(x1)\ln (x-1) is.
    Solution
    log(x1)\log(x-1)
    where (x1)>0 (x-1)>0
    x>1x>1
    Domain of log(x1)\log(x-1)  =(1,)=(1,\infty)
  • Question 3
    1 / -0
    The domain of the function, f(x)=log10(x4+6x)f(x) = \log_{10} (\sqrt{x-4}+\sqrt {6-x}) is
    Solution
    Given,
    f(x)=log10(x4+6x)f(x) = \log_{10} (\sqrt{x-4}+\sqrt {6-x})

    For given function to be defined,
    x4 0\Rightarrow x-4 \geq  0 \, and 6x 0 x 4\, 6-x \geq  0 \Rightarrow  x \geq  4 \, and x 6 \,x \leq  6

    \therefore  Domain of (x)=[4,6](x) = [4, 6]
  • Question 4
    1 / -0
    If f:[0,Π] [1,1]f : [0, \Pi ] \rightarrow  [-1, 1], f(x) = cosx, then f is.
    Solution
    f takes all values from [1,1][-1,1] while travelling from [0,π][0,\pi]. None of the values are repeated either. 

    Hence it is one-one and onto.
  • Question 5
    1 / -0
    If f(x)={1xQ0x Qf(x) = \left\{\begin{matrix} 1&x \in Q \\ 0 &x \notin  Q\end{matrix}\right. then fof(3)fof(\sqrt 3 ) is equal to
    Solution
    We know 3\sqrt{3} is an irrational number

    Thus f(3)=0f(\sqrt{3}) =0 and 00 is a rational number

    Hence fof(3)=f(0)=1fof(\sqrt 3 )= f(0) = 1
  • Question 6
    1 / -0
    The domain of the function log 3x2\log \displaystyle \sqrt {\frac {3-x}{2}} is.
    Solution
    We have f(x)=log3x2f(x) = \log \sqrt {\dfrac {3-x}{2}}

      3x2>0\because     \dfrac {3-x}{2}>0

     3x>0\Rightarrow 3 -x > 0

     x<3\Rightarrow x < 3

     x(,3)\Rightarrow x \in (-\infty , 3)

    \therefore   domain of f(x)f(x) is (,3)(-\infty , 3).
  • Question 7
    1 / -0
    The domain of the function f(x)=1x[x]  f(x) = \displaystyle \cfrac { 1 }{ \sqrt { x-\left[ x \right]  }  }
    Solution
    We have f(x)=1xxf(x) =\dfrac {1} {\sqrt{x-|x|}}

     x[x]>0\Rightarrow   x -[x] > 0

     x>[x]\Rightarrow   x > [x] is defined only when x Z.x\notin   Z.

    Domain of f(x)f(x) is RZR-Z
  • Question 8
    1 / -0
    The domain of the function f(x)=1(x1)(x2)\displaystyle f(x) =\frac {1}{\sqrt {(x-1) (x-2)}} is.
    Solution
    For f(x)f(x) to be defined (x1)(x2)>0(x-1)(x-2) > 0
    x(,1)(2,)\Rightarrow x \in (-\infty, 1) \cup (2, \infty)
  • Question 9
    1 / -0
    The domain of f(x)=logelogexf(x) = \log_e |\log_ex| is.
    Solution
    f(x)=loglogx,(x)f(x) = log|log x|, (x) is defined if logx>0|log x| > 0 and x>0x > 0 

    i.e.,ifx>0i.e., if x > 0 and xx \neq 1

    (logx>0  if   x 1)(\because |log x| > 0\,\, if\, \,  x \neq  1)

     x(0,1) (1,)\Rightarrow  x \in (0, 1) \cup  (1, \infty ).
  • Question 10
    1 / -0
    If functions f(x)f\left ( x \right ) and g(x)g\left ( x \right ) are defined on RRR\rightarrow R such that
    f(x)=x+3,xf(x)=x+3, x  \in   rational
             =4x,x =4x, x \in irrational
    g(x)=x+5g(x)=x+\sqrt{5}, x\in irrational
           =x,x  =-x, x \in rational
    then (fg)(x)\left ( f-g \right )\left ( x \right ) is
    Solution
    f(x)=x+3,xϵf\left( x \right) =x+3, x\epsilon  rational 
              =4x,xϵ=4x, x\epsilon  irrational

    g(x)=x+5,xϵg\left( x \right) =x+\sqrt { 5 } , x \epsilon  irrational
              =x,xϵ= -x,x\epsilon  rational 

    (fg)(x)=2x+3,x ϵ(f-g)(x)=2x+3, x\ \epsilon  rational
                         +3x5 xϵ+3x-\sqrt { 5\quad  } x\quad \epsilon irrational

    For one-one; 
    We know that one one function is a fn { f }^{ \underline { n }  } for which every element of the range of the function corresponds to exactly one element of the domain.
     
    But in this case this is not true as for all rational nos. thefn  { f }^{ \underline { n }  } is one one but for irrational fn { f}^{ \underline { n }  }, every elements of the range does not corresponds to exactly one element of the domain. 

    (fg)(x) (fg)(x )(f-g)(x)\neq  (f-g)({ x }^{ \prime  })         when xϵx \epsilon rational and 
                                                                   xϵx \epsilon irrational
    for onto:
    The fn (fg)(x){ f }^{ \underline { n }  }(f-g)(x) does not cover the whole range of the function.
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